Two radiations having energies E_1 and E_2 as | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$.For the first radiation: $E_1 = \frac{hc}{\lambda_1} = 15 \ eV$ $(I)$Fo

Vedclass · Accepted Answer

$\lambda_1 = 3\lambda_2$

Vedclass · Answer

(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$.For the first radiation: $E_1 = \frac{hc}{\lambda_1} = 15 \ eV$ $(I)$For the second radiation: $E_2 = \frac{hc}{\lambda_2} = 45 \ eV$ $(II)$Dividing equation $(I)$ by equation $(II)$:$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = \frac{15}{45} = \frac{1}{3}$Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{1}{3}$,which implies $\lambda_1 = 3\lambda_2$.

Two radiations having energies $E_1$ and $E_2$ as $15 \ eV$ and $45 \ eV$ respectively,then the relationship between $\lambda_1$ and $\lambda_2$ will be:

Similar Questions

The wavelength of sodium light is $5.862 \times 10^{-16} \, m$. The energy of the photon will be .....

$A$ gas absorbs a photon of $355 \ nm$ and emits at two wavelengths. If one of the emissions is at $680 \ nm,$ the other is at : ................. $nm$

The photoelectric effect shows:

While passing through an applied magnetic field,which of the following undergoes the maximum deflection?

Calculate the wavelength of yellow light having a frequency of $5.09 \times 10^{14} \ Hz$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module